Introduction To Redox
Types of Reaction
 Oxidation-Reduction called Redox
Ionic compounds are formed through
the transfer of electrons.
An Oxidation-reduction reaction involves
the transfer of electrons.
We need a way of keeping track.
Oxidation States


A way of keeping track of the electrons.
Not necessarily true of what is in nature, but
it works.
need the rules for assigning (memorize).
The oxidation state of elements in their
standard states is zero.
Oxidation state for monoatomic ions are the
same as their charge.
states state of  Oxygen isOxidation
assigned an oxidation
2 in its covalent compounds except as a
peroxide.
 In compounds with nonmetals hydrogen is
assigned the oxidation state +1.
 In its compounds fluorine is always –1.
 The sum of the oxidation states must be
zero in compounds or equal the charge of
the ion.
Oxidation States
Assign the oxidation states to each
element in the following.
CO2
NO3H2SO4
Fe2O3
Fe3O4
Oxidation-Reduction
Transfer
electrons, so the oxidation
states change.
Na + 2Cl2  2NaCl
CH4 + 2O2  CO2 + 2H2O
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
OIL RIG
LEO GER
Oxidation-Reduction
Oxidation means an increase in
oxidation state - lose electrons.
Reduction means a decrease in
oxidation state - gain electrons.
The substance that is oxidized is called
the reducing agent.
The substance that is reduced is called
the oxidizing agent.
Redox Reactions
Agents
Oxidizing agent gets reduced.
Gains electrons.
More negative oxidation state.
Reducing agent gets oxidized.
Loses electrons.
More positive oxidation state.
Identify the
Oxidizing agent
Reducing agent
Substance oxidized
Substance reduced
in the following reactions
Fe (s) + O2(g)  Fe2O3(s)
Fe2O3(s)+ 3 CO(g)  2 Fe(l) + 3 CO2(g)
SO3- + H+ + MnO4-  SO4- + H2O + Mn+2
Half-Reactions
All redox reactions can be thought of as
happening in two halves.
One produces electrons - Oxidation half.
The other requires electrons - Reduction half.
Write the half reactions for the following.
Na + Cl2  Na+ + ClSO3- + H+ + MnO4-  SO4- + H2O + Mn+2
Balancing Redox Equations



In aqueous solutions the key is the number of
electrons produced must be the same as
those required.
For reactions in acidic solution an 8 step
procedure.
Write separate half reactions
For each half reaction balance all reactants
except H and O
Balance O using H2O
Acidic Solution
 Balance H using H+
 Balance charge using e Multiply equations to make electrons
equal
 Add equations and cancel identical
species
 Check that charges and elements are
balanced.
Practice
The following reactions occur in aqueous
solution. Balance them
Cr(OH)3 + OCl- + OH- 
-2
CrO4 + Cl + H2O
MnO4- + Fe+2 Mn+2 + Fe+3
Cu + NO3-  Cu+2 + NO(g)
Pb + PbO2 + SO4-2  PbSO4
Mn+2 + NaBiO3  Bi+3 + MnO4-
Now
for
a
tough
one
-4
Fe(CN)
+ MnO 
6
NO3-
4
Mn+2 + Fe+3 + CO2 +
Basic Solution
Do everything you would with acid, but
add one more step.
Add enough OH- to both sides to
neutralize the H+
CrI3 + Cl2  CrO4- + IO4- + ClFe(OH) + H O  Fe(OH)
2
2 2
Redox Titrations
Same as any other titration.
the permanganate ion is used often because
it is its own indicator. MnO4- is purple, Mn+2
is colorless. When reaction solution remains
clear, MnO4- is gone.
Chromate ion is also useful, but color change,
orangish yellow to green, is harder to detect.
Example
The iron content of iron ore can be
determined by titration with standard
KMnO4 solution. The iron ore is dissolved
in excess HCl, and the iron reduced to
Fe+2 ions. This solution is then titrated
with KMnO4 solution, producing Fe+3 and
Mn+2 ions in acidic solution. If it requires
41.95 mL of 0.205 M KMnO4 to titrate a
solution made with 0.6128 g of iron ore,
what percent of the ore was iron?